Question

Light of wavelength 121.6 nm is emitted by a hydrogen atom. What are the (a) higher quantum number and (b) lower quantum number of the transition producing this emission? (c) What is the series that includes the transition?

Short answer

(a) Higher quantum number, ${\mathrm{n}}_1=2$

(b) Lower quantum number,${\mathrm{n}}_2=1$

(c) Lyman series.

Step-by-step solution

Identification of the given data:

The given data is listed below.

Wavelength of light is $\mathrm{\lambda }=121.6\mathrm{nm}$.

The principal Quantum number:

The principal quantum number is used to describe the electron’s state and is the one-four quantum number assigned to each electron in an atom.

The value of the principal quantum number is a natural number.

(a) Determine the value of the higher quantum number:

The energy emitted by the hydrogen atom of the photon is expressed as,

$E_{ph}=A\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

Now, multiply both sides by $\frac{1}{hc}$, and you have

$\frac{E_{ph}}{hc}=\frac{A}{hc}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

Also as the wavelength is,

$\frac{1}{\mathrm{\lambda }}=\frac{{\mathrm{E}}_{\mathrm{Ph}}}{\mathrm{hc}}$

Substitute$\frac{A}{hc}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$for$\frac{{\mathrm{E}}_{\mathrm{P}}\mathrm{h}}{\mathrm{hc}}$in the above equation.

$$\begin{gathered} \frac{1}{\lambda }=\frac{A}{hc}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \\ \frac{hc}{\lambda A}=\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \end{gathered}$$

Substitute 121.6 nm for $\lambda$, $6.26\times {10}^{-34}\mathrm{J}.\mathrm{s}$for h , $3\times {10}^8\mathrm{m}/\mathrm{s}$for c , and 13.6 eV for A in the above equation.

$$\begin{gathered} \frac{\left(6.26\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{\left(121.6\times {10}^{-9}\mathrm{m}\right)\left(13.6\times 1.602\times {10}^{-9}\mathrm{J}\right)}=\left(\frac{1}{{\mathrm{n}}_2^2}-\frac{1}{{\mathrm{n}}_1^2}\right) \\ 0.75=\left(\frac{1}{{\mathrm{n}}_2^2}-\frac{1}{{\mathrm{n}}_1^2}\right) \end{gathered}$$

To satisfy the above equation, the values of ${\mathrm{n}}_2$and ${\mathrm{n}}_1$are 1 and 2 respectively. Thus,

$\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}=0.75$

Thus, the higher quantum number is ${\mathrm{n}}_1=2$.

(b) Determine the value of lower quantum number:

And so,

$\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}=0.75$

Thus, the lower quantum number is $n_2=1$.

(c) Determine the name of the series that includes the transition:

The name of the series that includes the transition is Lyman series transition as this transition occurs when the electron goes from $n_1$to $n_2$emitting a photon.